Find the value of the following limit:
\huge\lim_{x\to\infty}e^{e^{e^{\biggl(x\,+\,e^{-\left(a+x+e^{\Large x}+e^{\Large e^x}\right)}\biggr)}}}-e^{e^{e^{x}}}I don’t even know how to start with. (this problem was shared in Brilliant.org)
Some of the ideas I tried is to take the natural log of this expression and reduce it to \ln(a/b) then use L’Hopital’s but that made it false!!
I know the value of the limit it is e^{-a} but please how to prove it?
Answer
Let’s observe that the limit is of form
f\left(x+\frac{\alpha}{f'(x)}\right) – f\left(x\right)
where f(x)=e^{e^{e^x}} and \alpha=e^{-a}. Since f is differentiable, we have
f\left(x+\frac{\alpha}{f'(x)}\right) – f\left(x\right)=
\frac{\alpha}{f'(x)}f’\left(x+\xi\right)
for some 0\leq \xi \leq \frac{\alpha}{f'(x)}. It remains to prove that f is “continuous enough” for this limit to converge as it seems it should, that is
\frac{f’\left(x+\xi\right)}{f’\left(x\right)}\to 1
I’ll try to find some clever way to do this.
Edit: Well, brutal force will do. We have
\xi\leq \alpha\, e^{-e^{e^x}}e^{-e^x}e^{-x}
Now,
\frac{f’\left(x+\xi\right)}{f’\left(x\right)}=
e^{e^{e^{x+\xi}}-e^{e^x}} e^{e^{x+\xi}-e^x}e^\xi
Obviously, e^\xi\to 1. Furthermore, by above inequality,
e^{x+\xi}-e^x=e^x\left(e^\xi – 1\right)=e^xO\left(\xi\right)\to 0
and
e^{e^{x+\xi}}-e^{e^x}=e^{e^x}\left(e^{e^{x+\xi} -e^x}-1\right)=
e^{e^x}O\left(e^{x+\xi} -e^x\right)=e^{e^x}e^x O\left(\xi\right)\to 0
That finishes the proof.
Attribution
Source : Link , Question Author : user153330 , Answer Author : Marcin Łoś
